Let $0<R<\pi/2$ be fixed. We want to show that for all sufficiently large $n\in \mathbb{N}$, the polynomial $$ P_n(z):=1+\frac{z^2}{2!}+\frac{z^4}{4!}+\ldots+\frac{z^{2n}}{(2n)!}$$ has no roots with absolute value less than $R$.
My instinct when I'm given a function and asked to show there are x many roots in a certain region (here the disk of radius $R$) is to use Rouché's Theorem. I haven't found a way to make that work here.
Alternatively, there may be something simple I'm overlooking. I know that $$\cos(x)=\sum_{n=0}^\infty \frac{(-1)^nx^{2n}}{(2n)!}.$$ If it weren't for the $(-1)^n$ term... Being given that $0<R<\pi/2$ makes me want to use something more like this (trig. functions).
Any hints to get the ball rolling would be appreciated!